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प्रश्न
The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is ______.
पर्याय
126
0
135
160
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उत्तर
The smallest value of the polynomial x3 – 18x2 + 96x in [0, 9] is 0.
Explanation:
Let f(x) = x3 – 18x2 + 96x
So, f'(x) = 3x2 – 36x + 96
For local maxima and local minima f'(x) = 0
∴ 3x2 – 36x + 96 = 0
⇒ x2 – 12x + 32 = 0
⇒ x2 – 8x – 4x + 32 = 0
⇒ x(x – 8) – 4(x – 8) = 0
⇒ (x – 8)(x – 4) = 0
∴ x = 8, 4 ∈ [0, 9]
So, x = 4, 8 are the points of local maxima and local minima.
Now we will calculate the absolute maxima or absolute minima at x = 0, 4, 8, 9
∴ f(x)= x3 – 18x2 + 96x
`"f"(x)_(x = 0)` = 0 – 0 + 0 = 0
`"f"(x)_(x = 4)` = (4)3 – 18(4)2 + 96(4)
= 64 – 288 + 384
= 448 – 288
= 160
`"f"(x)_(x = 8)` = (8)3 – 18(8)2 + 96(8)
= 512 – 1152 + 768
= 1280 – 1152
= 128
`"f"(x)_(x = 9)` = (9)3 – 18(9)2 + 96(9)
= 729 – 1458 + 864
= 1593 – 1458
= 135
So, the absolute minimum value of f is 0 at x = 0
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